Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The way I interpret the resistance \$R\$ of a resistor, which has dimensions \$ [\frac{\mathrm{V}}{\mathrm{A}}] \$ is: how many volts must be applied across the resistor to achieve 1 ampere of current?

The conductance \$G\$, which has dimensions \$ [\frac{\mathrm{A}}{\mathrm{V}}] \$ is then: how many amperes of current flow through the resistor when applying 1 volt?

I realize that these quantities are related to the geometry, whereas the resistivity \$ \rho = R\times \frac{A}{L} \$ which has dimensions \$ [\Omega \cdot \mathrm{m}] \$ is an intrinsic property of the material (doping of semi-conductor, electron/hole mobility, etc).

However, I cannot achieve an intuitive understanding to interpret the dimensions of resistivity. Can this be clarified?

share|improve this question
up vote 8 down vote accepted

\$\Omega\$m is the simplified unit of resistivity. The full unit is \$\Omega \$m\$^2/\$m. This means that a given length of material with a given cross sectional area will have a certain resistance whose value can be calculated using the resistivity.

For a 1 m length of material with a 1 mm\$^2\$ cross sectional area and a resistivity of 1:

\$1 \Omega \mathrm{m} = R(10^{-6}\mathrm{m}^2/1\mathrm{m})\$

\$R = {1\Omega \mathrm{m} \over(10^{-6}\mathrm{m}^2/1\mathrm{m})} = 10^6\Omega\$

share|improve this answer

Maybe it's easier to understand intuitively if you don't reduce the dimensions, in the same way of thinking as the units of gain as volts/volts, you could think of the units of resistivity as \$\frac {\Omega \mathrm{m}^2}{\mathrm{m}}\$, which fits with the physical interpretation of a resistive object of constant cross-sectional area and a given length.

Consider also the usual dimension for sheet resistivity, which is \$\Omega\$ per square, where "per square" really doesn't mean anything dimensionally, but prevents confusion with simply saying \$\Omega\$.

Another example, the units of torque (\$\mathrm{n}\cdot \mathrm{m}\$) are the same as the units of work- it's the physical interpretation that makes the difference.

share|improve this answer

The simplified unit Ω·m can be interpreted as:

If you multiply it by a length — specifically a thickness — then you obtain the resistance of an arbitrarily-sized square of that material with that thickness.

This is less useful in practical situations than the un-simplified interpretation.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.